# Complex tetrad vs. Real metric

I asked this question almost a month ago on mathoverflow (https://mathoverflow.net/q/228138/) but received no response. I thought I could have better luck here:

I have a question on the relationship between the complex tetrad in general relativity and the metric. All the papers I've sen so far just usually state the metric and the (null) tetrad without discussing the relation between the two.

My question is: clearly, null tetrad can be complex. Some of these complex tetrads do give a real metric, but not all. For example, the Kinnersley tetrad (coordinate system ($t,s,x,y$))

$$l=(0,1,0,0)$$

$$n=\rho\overline{\rho}(s^{2}+a^{2},-\frac{1}{2}(s^{2}-2ms+a^{2}),0,a)$$

$$m=-\frac{1}{\sqrt{2}}\overline{\rho}(ia\sin x,0,1,i\csc x)$$

where $\rho=-\frac{1}{s-ia\cos x}$, is supposed to give the Kerr metric. The Kerr metric is of course real. But when I substitute this null tetrad directly, I get a metric with complex entries. Just simply taking the real part of it doesn't do the trick, i.e. I don't get Kerr.

Attempt at the solution:

I first calculate the 1-forms corresponding to the tetrad above as

So how do I get a real metric from a complex tetrad?

• With your notations the tetrad is $\{l, n, m, \overline{m}\}$, then the metric is $g=-l\otimes n - n\otimes l + m\otimes\overline{m}+\overline{m}\otimes m$, which is real. – MBN Mar 10 '16 at 16:54